MATHEMATICAL METHODS

Transcription

1 8 Practice Exam A Letter STUDENT NUMBER MATHEMATICAL METHODS Writte examiatio Sectio Readig time: 5 miutes Writig time: hours WORKED SOLUTIONS Number of questios Structure of book Number of questios to be aswered Number of marks A B 5 5 Total 8 Studets are permitted to brig ito the examiatio room: pes, pecils, highlighters, erasers, sharpeers, rulers, a protractor, set squares, aids for curve sketchig, oe boud referece, oe approved techology (calculator or software) ad, if desired, oe scietific calculator. Calculator memory DOES NOT eed to be cleared. For approved computer-based CAS, full fuctioality may be used. Studets are NOT permitted to brig ito the examiatio room: blak sheets of paper ad/or correctio fluid/tape. Materials supplied Questio ad aswer booklet of 5 pages. Formula sheet. Workig space is provided throughout the book. Istructios Write your studet umber i the space provided above o this page. Uless otherwise idicated, the diagrams i this book are ot draw to scale. All writte resposes must be i Eglish. At the ed of the examiatio You may keep the formula sheet. Studets are NOT permitted to brig mobile phoes ad/or ay other uauthorised electroic devices ito the examiatio room. TRIUMPH TUTORING 7

2 8 MATHMETH EXAM A - TRIUMPH TUTORING WORKED SOLUTIONS 8 Exam A - Sectio A Questio Period of taget fuctio = b P = P = Aswer: A Video Solutio: Questio f(x) =e x+ x g(x) =log e g(f(x)) = log e (ex+ ) g(f(x)) =log e (e x+ ) g(f(x)) =log e (e) g(f(x)) =(x +) g(f(x)) = x Questio Aswer: B Origial = P (, ) Up = P (, ) Left = P (, ) Reflected = P (, ) ) Fial = P (, ) Aswer: E Video Solutio: Questio 5 b a Video Solutio: Questio If f ( ) =, the gradiet is at x =. If f (x) > for x R\{}, the gradiet is positive everywhere else. This meas you have a positive cubic which has a statioary poit of iflectio at x =. Aswer: C Video Solutio: From diagram, a = b. Pr( X Z <Z<) = Pr(.9 <X<.9) Pr( <Z<) =Pr( <Z<) Pr( <Z<) = Pr(.9 <X<.9) Aswer: A Video Solutio:

3 8 MATHMETH EXAM A - TRIUMPH TUTORING Questio Use CAS: a< p or a> + p f(x) is the derivative of chose graph (A). A has a turig poit at x =meaig f(x) has a x-itercept at x =. The gradiet of A is egative util x =, the positive from x>. Aswer: A Video Solutio: x g(x) =log e g(x) =log e (a) g(x) =log e x g(x) =log e (x) g(x) =log e ((x) ) g(x) =log e (8x ) ) a =8x Aswer: D Questio 7 Video Solutio: Questio 8 If graphs itersect, let them equal each other: ax =x x x x ax += x +( a)x += For two poits of itersectio, >. ) b ac > ( a) ( ) > + a +a > a +a 8 > (a +a ) > a +a > Aswer: D Video Solutio: Questio 9 If the ormal has a gradiet of has a gradiet of. ) f () =. Let y = ae u, where u = x : dx = du du dx dx = aeu x dx =axex dx =at x = =a()e () =ea a = e Aswer: E, the taget Video Solutio: f(x) =x +x + d Questio If it has three distict solutios, the two turig poits are o either side of the x-axis. We eed to fid where the turig poits are, so f (x) =. f (x) =x +x = =x(x +) x =,x= Let s preted d =for a secod.

4 8 MATHMETH EXAM A - TRIUMPH TUTORING f(x) =x +x f() = f( ) = ( ) +( ) f( ) = 8 Turig poits at (, ), (, 8) I order for there to be three x-itercepts, the graph eeds to be moved dow by at most 8 uits. Therefore, sice we re addig d, d must be betwee 8 ad. Aswer: E (, 8) y (, ) Video Solutio: Questio Drawig a Ve Diagram is the easiest way to visualise this. A Sice Pr(A [ B) = 8 9 ad x 5 x Pr(A \ B )=Pr(A \ B), x x = 8 9 B x x = 5 x = 9 Pr(A \ B) Pr(A B) = Pr(B) 5 Pr(A B) = Pr(A B) = 5 a 9 9 a Pr(A B) = 8 9 Aswer: D Video Solutio: Questio f(x) =a cos(b(x + c)) + d Period = b 5 = b b = 5 Midpoit of rage = + Midpoit of rage = Distace betwee midpoit ad max: = = 7 ) Amplitude = 7 x = Horizotal traslatio =

5 5 8 MATHMETH EXAM A - TRIUMPH TUTORING ) f(x) = 7 cos 5 Aswer: D x Video Solutio: Approximatio: Questio Area = A + A + A + A Area = + Area = + + Area = Area =5 Exact: Area = Z 5 Area = 5 (CAS) % differece = % differece = % differece = x dx approx exact % differece = 8 % % Aswer: D Video Solutio: Questio Cofidece Iterval =(5%, %) Cofidece Iterval =(.5,.) r r ˆp( ˆp) ˆp( CI = ˆp z, ˆp + z! ˆp) ).5 =.55.9 Solve for i CAS: =7 Aswer: C r.55(.55) Video Solutio: x x x x x x = x +5 x Questio 5 = ( x)+5 x = ( x) ( x) x x = + 5 x + 5 x ) Asymptotes at x =ad y = Alteratively, sketch this o your CAS ad use that to fid asymptotes. Aswer: D Video Solutio: Turig poits at x = Questio ad x =, ad domai must ot go past turig poits to allow for a oe-to-oe iverse fuctio. The oly optio that works with [b, ) is D. Sketch the graph o CAS to visualise. Aswer: D Video Solutio: r ˆp( ).5 = ˆp z ˆp) ˆp =.55 (halfway betwee.5 ad.) z =.9 (kow for 95% CI)

6 8 MATHMETH EXAM A - TRIUMPH TUTORING Questio 7 We eed two simultaeous equatios to solve for a ad b. [] : Area uder graph = ) = Z b [] : E(X) = ) Zb = ax dx [] x ax dx [] Usig Solve Systems of Equatios i CAS: a = 8 8 ad b = Aswer: C Video Solutio: Questio 8 B = Takes a black coi W = Takes a white coi +k k +k B W +k k +k +k k +k Pr(BB) = 7 ) 7 = +k +k B W B W Solve i CAS: k = 9 or k =. Sice k>, k = Aswer: C Video Solutio: Chai rule: y = e u Z u = log e (x)+dx Questio 9 u = x log e (x)+c (usig our CAS) Passes through (, log e (7)): log e (7) = log e () + c log e (7) = log e (7) + c c = ) u = x log e (x) Whe u =log e (): log e () = x log e (x) log e () = log e (x x ) =x x ) x = Usig our chai rule: dx = du du dx dx =eu (log e (x)+) Method : Substitute u = x log e (x) ) dx =ex log e (x) (log e (x)+) ) dx =elog e (xx ) (loge (x)+) ) dx =xx (log e (x)+) Whe x =: dx =()() (log e () + )

7 7 8 MATHMETH EXAM A - TRIUMPH TUTORING dx = (log e () + ) dx =log e () + Aswer: A Method : Whe u =log e (), x =: dx =elog e () (log e () + ) dx =elog e ( ) (loge () + ) dx =() (log e () + ) dx =log e () + Aswer: A Video Solutio: Questio Volume = Legth Width Height where L = x where W = x where H = x V =( x)( x)x V =x x +x To fid where V is a maximum, V (x) = V (x) =x x + Accordig to CAS: V (x) =whe x =.or x =. Sice x <, x< ) x =. Aswer: B Video Solutio:

8 8 8 MATHMETH EXAM A - TRIUMPH TUTORING WORKED SOLUTIONS 8 Exam A - Sectio B h(x) =a + b si(cx) Questio a Sice maximum height =8m, We use a b = 8, because b is egative (accordig to graph). At poit B, h(x) =. Sice the track is smooth here, this must be the miimum of h(x). If miimum height =m, We use a + b =, because b is egative (accordig to graph). ) We have two simultaeous equatios: a b =8 [] a + b = [] From [], a = b +8 Sub ito []: (b +8)+b = b +8= b = 8 ( st mark = c c = ( rd mark a =,b= 8, c= Video Solutio: Questio b Max height =8metres x 8 = 8si x =si si ( ) = x = x ) x =8metres at maximum height * st mark Video Solutio: Sice b = 8, a = 8+8 a =( d mark Period = c Questio c We kow that from A to B is metres. This is ad periods of h(x). To fid what oe period is: P 5 = P = 5 P =metres P = c Period = Period =metres ( st mark Video Solutio: Questio d x 5 = 8si 5 x 8 =si

9 9 8 MATHMETH EXAM A - TRIUMPH TUTORING si 5 8 x = x =.75,.87, 5.8 ( st mark x =.58,.58,. Sice x>, first time h reaches 5 metres is at: x =.58 ( d mark Video Solutio: Questio e Coaster is oly above 5 metres high betwee x =.58 ad x =.. So, above 5 metres for (..58) horizotal distace. Horizotal distace travelled while above 5 metres =.8 metres ( st mark Distace travelled below 5 metres: =.8 =. metres ( d mark Video Solutio: Questio a Starts fallig at turig poit: TP =(, ) h(x) =a(x ) +( st mark Emily is metres away, so x-it =(, ) =a( ) + = a() a = 7 ) h(x) = 7 (x ) +( d mark Video Solutio: Questio b 7 h() = ( ) + 7 h() = + h() = 7 + h() = metres ( st mark Video Solutio: Questio c Draw triagle to figure out agle. A = agle of depressio Due to parallel lies, at A = at B. ta( ) = =ta B * st mark =9.9 ( d mark Video Solutio: Questio di Create trasformatio matrix: x 5A = y x = x y = y + 5 x 5 + y 5 * st mark

10 8 MATHMETH EXAM A - TRIUMPH TUTORING x = x y =y Sub x ad y ito h(x) for g(x): y = 7 x + y = 7 x + 9 y = 7 x + 9 ) g(x) = 7 x + 9 * d mark * rd mark Video Solutio: Questio dii Need to fid x-itercept of g(x) x-it whe g(x) = 7 = x = 7 x 9 7 = x 7 = x x = (±p 7 + ) x = ( p 7) ad x = ( + p 7) Sice x>, x = ( + p 7) x =. ( st mark (.) = 5.9 Decreased by 5.9 metres ( d mark Video Solutio: Questio diii Fid y-value of turig poit: g(x) = 9 metres ( st mark Video Solutio: E(X) =mea = E(X) = Z Questio a Z b a x x 9 x f(x) dx 5x +x dx ) E(X) =. hours ( st mark Video Solutio: = Z. = 9.87 = ) = x 9 apple x Questio b 5x 5x +x dx +x ( st mark 5() +() Accordig to CAS: =. or.5 Sice apple apple, =.5 hours ( d mark Video Solutio: Questio ci = ( st mark 8 Video Solutio:

11 8 MATHMETH EXAM A - TRIUMPH TUTORING Questio cii Questio di G = Goes to the gym G = Does t go to the gym Use Biomial Pdf i CAS: =,p= 7,x= WED G THUR G G FRI G G G G * st mark ) Pr(X =)=.57 ( st mark Video Solutio: Questio dii Use Biomial Cdf i CAS: Pr(X >5) = Bi X, 7 ( st mark lower =, upper =,=,p= 7 Pr(G )=Pr(GGG )+Pr(GG G ) Pr(G ) = + Pr(G ) = + 8 Pr(G )= ( d mark 8 Video Solutio: Questio ciii Pr(G Thur G Fri) = Pr(G Thur \ G Fri) Pr(G Fri) Pr(G Thur G Fri) = Pr(G Thur G Fri) = 8 a a 8 * st mark Pr(G Thur G Fri) = ( d mark Video Solutio: ) Pr(X >5) =.7 ( d mark Video Solutio: Questio ei ˆp apple = X apple ˆp apple =X apple ˆp = X ˆp =X Pr ˆp apple ˆp =Pr(X apple X ) Pr(X apple X ) = Pr( apple X apple ) Pr(X ) Use Biomial Cdf o CAS to fid Pr( apple X apple ): * st mark lower =, upper =,=,p= 7 ) Pr( apple X apple ) =.5

12 8 MATHMETH EXAM A - TRIUMPH TUTORING Use Biomial Cdf o CAS to fid Pr(X ): lower =, upper =,=,p= 7 ) Pr(X ) =.9979 ( d mark ) Pr(X apple X ) = Questio eii r r! ˆp( ˆp) ˆp( ˆp) Cofidece Iterval = ˆp z, ˆp + z v u t v u t CI = +.9 C A ) Pr(X apple X ) =.5 Pr ˆp apple ˆp =.5 * rd mark Video Solutio: CI =(.99,.5) ( st mark Video Solutio: Questio ai For poits of itersectio, let h(x) =g(x). x = x +x + t = x +x + t ( st mark To solve, use quadratic formula, where: a =,b=,c= t x = b ± p b ac a q x = ± () t x = ± p +t x =± p +t ( d mark g( + p +t) = + p +t + t g( g( g( p +t) = p +t p +t) = p +t) = p +t ++t p +t + t p p ) A = +t, +t + t )B = + p +t, + p +t + t * rd mark Video Solutio: g( + p +t) = + p +t g( + p +t) =+ p +t ++t

13 8 MATHMETH EXAM A - TRIUMPH TUTORING Area = Area = Z b a h(x) + Z p +t p +t Questio aii g(x) dx apple x Area = + x + tx Area = x +x + t x dx + p +t + p +t p +t * st mark + + p +t + t + p +t + p +t! p +t + t p +t! Questio c A = p (t +) da dt = p (t +) ( st mark da dt =p p t + da dt =p +t ( d mark Video Solutio: log e () = p +t log e () = p +t +t = (log e ()) t = (log e ()) Questio d A = p (t +) uits t = (log e ()) 8 ( st mark * d mark Video Solutio: Video Solutio: At t =, Area = k ) k = p (+) k = p (8) k = p Questio b uits ( st mark Video Solutio: f(x) =ke x f() < ) >ke >k k< Questio 5a Sice k > (because the graph is a positive expoetial), <k< ( st mark Video Solutio:

14 8 MATHMETH EXAM A - TRIUMPH TUTORING P lies at (p, f(p)) Questio 5bi Sice f(x) =e x (after lettig k =) f(p) =e p ) P at (p, e p ) ( st mark Legth formula: q L = (x x ) +(y y ) q L = (p ) +(e p ) L = p p +(e p ) f(.5) = e.5 f(.5) =. ) P lies at (.5,.) ( rd mark Video Solutio: Questio 5biii L = p p + e p e p + q L(.5) = (.5) + e (.5) e.5 + L =.9 uits ( st mark L = p p + e p e p +( d mark Video Solutio: Video Solutio: Questio 5bii If OP is a miimum, L is a miimum. Whe L is a miimum, dl dp =. L = p p + e p e p + Let L = p u, where u = p + e p e p + dl dp = dl du du dp dl dp = p u (p +ep e p ) dl dp = p + e p e p p ( st mark p + e p e p + dl =at miimum: dp = p + e p e p p p + e p e p + Questio 5ci P occurs at (.5,.) f(x) =e x f (x) =e x f (.5) = e.5 f (.5) =.9 ( st mark For taget: y y = m(x x ) y (.) =.9(x.5) y =.9x.89. ) g(x) =.9x. ( d mark Video Solutio: Solve for p i CAS: p =.5 ( d mark

15 5 8 MATHMETH EXAM A - TRIUMPH TUTORING Area = A = A = Z Z log e () log e () Z log e () Questio 5cii f(x) g(x) dx (e x ) (.9x.) dx e x.9x.8 dx A =[e x.8x.8x] log e () ( st mark A = e log e ().8 (log e ()).8(log e ()) e.8 ().8() A =. uits ( d mark Video Solutio: P =(.5,.) Questio 5di Dilatio by a factor of. from the y-axis (multiply x-value by.): New poit =(.5.,.) New poit =(.,.) ( st mark Traslatio of.9 i positive directio of y-axis: Q =(.,. +.9) Q =(.,.) ( d mark Legth of OQ: Questio 5dii q L = (x q x ) +(y y ) L = x +(ke x ) Accordig to CAS: dl dx = k e x ke x + x p k e x ke x + x + dl dx =at miimum legth of OQ. Sice Q =(.,.), x =. at dl dx = ) = k e (.) ke (.) +(.) q k e (.) ke (.) +(.) + Multiply both sides by deomiator: =.5k.5k +. Accordig to CAS: k =.9 or k =.5 ( d mark ) a =.9 or a =.5 Sice a>, a =.5 ( rd mark * st mark Video Solutio: Video Solutio: